Researchers Stickgold, James, and Hobson investigated delayed effects of sleep deprivation on learning in a study published in Nature Neuroscience (2000). The goal of the study was to see whether the improvement scores tend to be higher for the unrestricted sleep treatment than for the sleep deprivation treatment.
Unrestricted-sleep group’s improvement scores (milliseconds):
Sleep-deprived group’s improvement scores (milliseconds):
\[ \bar{x}_\text{unrestricted} - \bar{x}_\text{deprived} = 15.92 \\[0.2in] \begin{align} H_0: \mu_\text{unrestricted} - \mu_\text{deprived} &= 0 \\ H_a: \mu_\text{unrestricted} - \mu_\text{deprived} &\neq 0 \end{align} \]
Because the null hypothesis asserts that improvement score is not associated with sleep condition, we will assume that the 21 subjects would have had exactly the same improvement scores as they did, regardless of which sleep condition group (unrestricted or deprived) the subject had been assigned.
Open the Multiple Means applet. Check the Show Shuffle Options box, select the Plot display, and press Shuffle Responses. Describe what happens when you press Shuffle Responses. How is the location of the next point on the dotplot determined?
Use a total of at least 1000 repetitions, and record a p-value. Record a sentence summarizing the strength of evidence in the context of this study.
Record an SE 95% confidence interval for \(\mu_\text{unrestricted} - \mu_\text{deprived}\). Is zero a plausible value?
The Central Limit Theorem applies to differences in means as well. Conditions:
Data on birth weight and mother smoking habit:
# A tibble: 3 × 4
habit xbar s n
<chr> <dbl> <dbl> <int>
1 nonsmoker 7.27 1.23 867
2 smoker 6.68 1.60 114
3 <NA> 7.05 1.91 19
T-statistic for independent means:
\[ T = \frac{(\bar{x}_1 - \bar{x}_2) - 0}{\sqrt{s_1^2/n_1 + s_2^2/n_2}} \] Degrees of freedom equals minimum of \(n_1-1\) and \(n_2-1\).
# A tibble: 2 × 4
habit xbar s n
<chr> <dbl> <dbl> <int>
1 nonsmoker 7.27 1.23 867
2 smoker 6.68 1.6 114
Compute a T-score and a P-value for this hypothesis test.
Write a sentence interpreting your conclusion in the context of the study.
Rows: 1,000
Columns: 13
$ fage <int> 34, 36, 37, NA, 32, 32, 37, 29, 30, 29, 30, 34, 28, 28, 32, 24, 29, 35, 25,…
$ mage <dbl> 34, 31, 36, 16, 31, 26, 36, 24, 32, 26, 34, 27, 22, 31, 25, 20, 32, 33, 26,…
$ mature <chr> "younger mom", "younger mom", "mature mom", "younger mom", "younger mom", "…
$ weeks <dbl> 37, 41, 37, 38, 36, 39, 36, 40, 39, 39, 42, 40, 40, 39, 34, 37, 38, 41, 44,…
$ premie <chr> "full term", "full term", "full term", "full term", "premie", "full term", …
$ visits <dbl> 14, 12, 10, NA, 12, 14, 10, 13, 15, 11, 14, 16, 20, 15, 20, 10, 12, 15, 16,…
$ gained <dbl> 28, 41, 28, 29, 48, 45, 20, 65, 25, 22, 40, 30, 31, NA, 25, 70, 22, 38, 20,…
$ weight <dbl> 6.96, 8.86, 7.51, 6.19, 6.75, 6.69, 6.13, 6.74, 8.94, 9.12, 8.91, 8.00, 7.2…
$ lowbirthweight <chr> "not low", "not low", "not low", "not low", "not low", "not low", "not low"…
$ sex <chr> "male", "female", "female", "male", "female", "female", "female", "male", "…
$ habit <chr> "nonsmoker", "nonsmoker", "nonsmoker", "nonsmoker", "nonsmoker", "nonsmoker…
$ marital <chr> "married", "married", "married", "not married", "married", "married", "marr…
$ whitemom <chr> "white", "white", "not white", "white", "white", "white", "white", "white",…
weight ~ habit
Welch Two Sample t-test
data: weight by habit
t = 3.8166, df = 131.31, p-value = 0.0002075
alternative hypothesis: true difference in means between group nonsmoker and group smoker is not equal to 0
95 percent confidence interval:
0.2854852 0.8998751
sample estimates:
mean in group nonsmoker mean in group smoker
7.269873 6.677193
We are 95% confident that the mean weight of babies born to nonsmokers is between 0.3 and 0.9 pounds more than the mean weight of babies born to smokers.